grouping array elements according to other array elements

Question

start with something like this:

array (
  0 => 
  array (
    'co' => '1',
    'lo' => 'aaa',
  ),
  1 => 
  array (
    'co' => '1',
    'lo' => 'bbb',
  ),
  2 => 
  array (
    'co' => '1',
    'lo' => 'ccc',
  ),
  3 => 
  array (
    'co' => '1',
    'lo' => 'aaa',
  ),
  4 => 
  array (
    'co' => '1',
    'lo' => 'bbb',
  )
)

Then group array elements with index 'lo', counting values of 'co' which accompany them, to get eventually something similar to:

array (
  aaa => 2,
  bbb => 2,
  ccc => 1
)

So, you want us to do your work for you? Please show us what have you tried so far. — Amal
– Amal, Commented Sep 29, 2013 at 19:51

shennan · Accepted Answer · 2013-09-29 20:08:05Z

1

I agree with the others about showing the code that you've tried so far. Having said that, this should work:

$array = array (
  array (
    'co' => '1',
    'lo' => 'aaa',
  ),
  array (
    'co' => '1',
    'lo' => 'bbb',
  ),
  array (
    'co' => '1',
    'lo' => 'ccc',
  ),
  array (
    'co' => '1',
    'lo' => 'aaa',
  ),
  array (
    'co' => '1',
    'lo' => 'bbb',
  )
);

$new_array = array();

foreach($array as $a){
  if(!array_key_exists($a['lo'], $new_array)){
    $new_array[$a['lo']] = intval($a['co']);
  }else{
    $new_array[$a['lo']] = $new_array[$a['lo']] + intval($a['co']);
  }
}

print_r($new_array);

We're cycling through the array, creating a new array key if one doesn't exist, and adding to the array key value with the number specified in 'co' if it already exists. You may want some additional checks to be sure that 'co' and 'lo' exist as array keys in the original array before trying to parse/add them to your new array.

As a side note, there is no need to specify the keys of your original array as numbers because arrays are automatically indexed. Note that I have removed these numbers when declaring the array.

edited Sep 29, 2013 at 20:08

answered Sep 29, 2013 at 20:02

shennan

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1 Comment

pknoe3lh Over a year ago

Also for the first statement ;-) $new_array[$a['lo']] = intval($a['co']);

skmail · Accepted Answer · 2013-09-29 19:59:10Z

0

Like this ?

$dataArray = array (
    0 =>
    array (
        'co' => '1',
        'lo' => 'aaa',
    ),
    1 =>
    array (
        'co' => '1',
        'lo' => 'bbb',
    ),
    2 =>
    array (
        'co' => '1',
        'lo' => 'ccc',
    ),
    3 =>
    array (
        'co' => '1',
        'lo' => 'aaa',
    ),
    4 =>
    array (
        'co' => '1',
        'lo' => 'bbb',
    )
);

$finalResults = array();

foreach($dataArray as $data){
    if(isset($finalResults[$data['lo']])){
        $finalResults[$data['lo']]++;
    }else{
        $finalResults[$data['lo']] = 1;
    }
}


print_r($finalResults);

answered Sep 29, 2013 at 19:59

skmail

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1 Comment

pknoe3lh Over a year ago

co is the count i think you should sum it up

pknoe3lh · Accepted Answer · 2013-09-29 20:01:27Z

0

Here you are ;-)

$in=array (
  0 => 
  array (
    'co' => '1',
    'lo' => 'aaa',
  ),
  1 => 
  array (
    'co' => '1',
    'lo' => 'bbb',
  ),
  2 => 
  array (
    'co' => '1',
    'lo' => 'ccc',
  ),
  3 => 
  array (
    'co' => '1',
    'lo' => 'aaa',
  ),
  4 => 
  array (
    'co' => '1',
    'lo' => 'bbb',
  )
);

$out= array();

for($i=0;$i<count($in);$i++)
{
    $out[$in[$i]['lo']]=0;
}

for($i=0;$i<count($in);$i++)
{
    $out[$in[$i]['lo']]+=$in[$i]['co'];
}

print_r($out);

answered Sep 29, 2013 at 20:01

pknoe3lh

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